HW 6, Question 1: Forces in a Rotating Frame
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HW 6, Question 1: Forces in a Rotating Frame Consider a solid platform, like a \merry-go round", that rotates with respect to an inertial frame of reference. It is oriented in the xy plane such that the axis of rotation is in the z-direction, i.e. ! = !z^, which we can take to point out of the page. Now, let us also consider a person walking on the rotating platform, such that this person is walking radially outward on the platform with respect to their (non-inertial) reference frame with velocity v = vr^. In what follows for Question 1, directions in the non-inertial reference frame (i.e. the frame of the walking person) are deﬁned in the following manner: the x0-direction points directly outward from the center of the merry-go round; the y0 points in the tangential direction of rotation in the plane of the platform; and the z0-direction points out of the page, and in the same direction as the z-direction for the inertial reference frame. (a) Identify all forces. Since the person's reference frame is non-inertial, we need to account for the various ﬁctitious forces given the description above, as well as any other external/reaction forces. The angular rate of rotation is constant, and so the Euler force Feul = m(!_ × r) = ~0. The signiﬁcant forces acting on the person are: 0 • the force due to gravity, Fg = mg = mg(−z^ ), 2 0 • the (ﬁctitious) centrifugal force, Fcen = m! × (! × r) = m! rx^ , 0 • the (ﬁctitious) Coriolis force, Fcor = 2mv × ! = 2m!v(−y^ ), • the normal force the person experiences from the rotating platform N = Nz^0, and • a friction force acting on the person, f, in the x0y0 plane. In order for the forces to balance, the friction force f must counteract all other forces acting on 0 0 0 the system. In other words, f has components in the x - and y -directions: f = fx(−x^ )+fyy^. If we add up all components and set all sum-of-force components equal to zero, we get: 2 fx = m! r; (1) fy = 2mv!; (2) N = mg: (3) (b) Find the point of slippage. The person with velocity v will not slip so long as the friction force balances the combination of centrifugal and Coriolis forces, and the friction force does not exceed a value at which 1 the ﬁctitious forces dominate. However, slippage will occur when f reaches its maximum value: fmax = µsN, where µs is the coeﬃcient of static friction. We can use this limit and Equations 1-3 to ﬁnd r as a function of all other known quantities: 2 2 2 f = fx + fy 2 2 2 = (m! rslip) + (2mv!) 1 = µ2(mg)2 =) r = pµ2g2 − 4v2!2 (4) s slip !2 s (b) Find value for rslip. −1 −1 We're asked to consider a scenario where ! = 1 rad s , v = 1 m s , and µs = 1=2. We can take g = 9:806 m s−2, the local value of acceleration due to gravity. Plugging all of these values into Equation 4, we ﬁnd that rslip ≈ 4:48 m (5) 2 HW 6, Question 2: Coriolis Force on a Moving Train Let's consider a train in France at a latitude θ = 45 degrees (i.e. in the northern hemisphere) that travels in the North direction at a constant speed v. As mentioned in the homework prompt, the center of mass is located at a height z0 and the two rails are separated by a distance w0. (a) Find ratio of forces exerted by the rails onto the wheels. We ultimately want to examine the forces experienced by the rails that the train travels on. In order to do so, we need to identify all forces acting on the rigid body: • the force due to gravity, Fg = mg, acting at the center of mass of the train, • the (ﬁctitious) centrifugal force, Fcen = m! × (! × r), acting at the center of mass of the train, • the (ﬁctitious) Coriolis force, Fcor = 2mv × !, since the train is moving, • the reaction force that the eastern rail exerts on the eastern wheel, f1, and • the reaction force that the western rail exerts on the western wheel, f2. In the case of Earth's rotation, the rotation rate j!j = ! = 7:3 × 10−5 rad/s is very small, so that terms that are proportional to !2 are negligible when compared to the Coriolis force, since the latter term is proportional to !. We therefore only need to consider the signiﬁcant forces acting on the system, which add to zero since the train keeps moving at a constant speed v: F ≈ mg + 2mv × ! + f1 + f2 = 0 (6) We can take components of Equation 6 in the vertical and horizontal directions in the local, non-inertial frame of reference. The direction of the Coriolis force can be determined since the deﬁnition of the coordinate system dictates that ! points in the direction of the axis of rotation, and that v points due North; the right hand rule requires that the vector v × ! points due East. It has a magnitude of jFcorj = Fcor = 2mv! sin θ. Since we neglect the −2 small contribution of Fcen, the local value of the gravitational acceleration g = 9:806 m s and it points straight down towards the center of the Earth. Figure ?? shows the various force vectors acting at their respective points in the rigid body of the train. With these directions, we can write the general y-component and z-component force equations, ΣFz = f1;z + f2;z − mg = 0 (7) ΣFy = Fcor − f1;y − f2;y = 0: (8) 3 One initial approximation that we can make is that that the eastern rail experiences a much larger horizontal reaction force that opposes the Coriolis term than the western rail. From Equation 8, we therefore require that the sum f1;x + f2;x ≈ f1;x; by extension, this approximately means that f2 points strictly in the vertical direction, since it has no signiﬁcant horizontal component and so f2 ≈ f2;yz^, while f1 has both horizontal and vertical components. If we make the substitution jf2j ≈ f2;y = f2, we can re-write the component equations as ΣFz ≈ f1;z + f2 − mg = 0 (9) ΣFy ≈ Fcor − f1;y = 0: (10) Equations 9 and 10 contain three unknowns, so we cannot uniquely solve for each compo- nent using these two equations alone. However, we can proceed further with a few more approximations using two slightly diﬀerent methods: Method 1: Equilibrium of Forces and Torques Since the train remains on the track and moves at a constant speed, the sum of all forces and of all torques must cancel to yield no net terms. We can add all moments of force and require that their sum be zero; in order to do so, as with calculations of angular momentum, we must select a reference point to measure moments from. In Figure ??, the reference point is chosen (arbitrarily) to be at the western wheel (where f2 acts). The sum of all moments relative to this point are: ΣM = (0× f2) + (rCOM × Fg) + (rCOM × Fcor) + (r2 × f1) x^ y^ z^ x^ y^ z^ x^ y^ z^ = 0 w0=2 z0 + 0 w0=2 z0 + 0 w0 0 0 0 −mg 0 Fcor 0 0 f1;y f1;z mgw =x ^ − 0 − z F + w f (11) 2 0 cor 0 1;z = 0 Using Equation 11, we can ﬁnd the value of f1;z in terms of the forces we can compute and in terms of the distances where the forces act: mg z0 f1;z = + Fcor (12) 2 w0 Method 2: Force Approximations Instead of invoking torques, we can make further approximations of the force components through physical arguments. Let's the consider the simplest, \everyday" case, where we ig- nore the non-inertial forces and simply consider the force due to gravity and the two reaction 4 forces on the rails. In the simplest case, the reaction forces are equal, and so f1 = f2 = mg=2 and each point in thez ^ direction, so that the sum of the two forces perfectly balance the weight of the train. In the case at hand, where we consider the Coriolis force to ﬁrst order in !, Equations 9 and 10 suggest that f1 and f2 are slightly modiﬁed. From Equation 10, we see that f1;y = f1 sin φ = Fcor, where tan φ = Fcor=Fg (see Figure 1). Given the assumptions of how the reaction forces are generated, f1;z should be equal to mg=2 with the addition of a term that is proportional to Fcor. From Equation 9, we see that the Coriolis term enters in the z-component Equation as f1 cos φ = (f1 sin φ) cot φ = Fcor cot φ. If we further assume that this component of the Coriolis term, as with the weight, is distributed evenly between the two rails, then we can write the ﬁrst-order equation for f2 as mg 1 f = − cot φF 2 2 2 cor since one would expect the eastward force to slightly lessen the reaction exerted by the western rail. For an order-of-magnitude calculation, we can require that tan φ ∼ z0=(w0=2) = 2z0=w0. Therefore, with the above equation for f2, we can ﬁnd the modiﬁed equation for f1;z using Equation 9: 1 f ≈ mg + cot φF 1;z 2 cor mg z0 = + Fcor (13) 2 w0 Conclusions The combination of Equations 9, 10 and 12 allows us to uniquely solve the system for the components we want. Since we're asked to ﬁnd the ratio of the magnitudes of rail forces, we must ﬁrst compute the magnitude of f1; we can do this by adding the components in quadrature and substituting the results from Equations 10 and 12: 2 2 2 f1 = f1;y + f1;z 2 2 mg z0 = Fcor + + Fcor 2 w0 2 2 2 z0 (mg) mgz0 = Fcor 1 + 2 + + Fcor w0 4 w0 2 (mg) mgz0 ≈ + Fcor (14) 4 w0 2 2 where, in the ﬁnal form of Equation 14, we note that Fcor / ! and drop this term since ! is very small.